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I've verified conjectured relationship (3) below for several thousand values of $x$ and suspect that it's true.

(1) $\quad H(x)=\sum\limits_{n=1}^x\frac{1}{n}\qquad\qquad\qquad\quad$(Harmonic Number Function)

(2) $\quad M(x)=\sum\limits_{n=1}^x\mu(n)\qquad\qquad\qquad$(Mertens Function)

(3) $\quad H(x)\stackrel{?}{=}\sum\limits_{n=1}^x\frac{\sigma_1(n)}{n}\,M\left(\frac{x}{n}\right)\qquad\quad(\sigma_1(n)$ is the Sum-of-Divisors Function$)$

Question: Has conjectured relationship (3) above been proven or disproven (or can it be)?

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  • $\begingroup$ I don't have a laid out proof, but something that seems like it could be useful in the proof of this is that $\sum_{n=1}^x M(\frac{x}{n}) = 1$ $\endgroup$ – Ryan Shesler Jun 23 '19 at 21:58
  • $\begingroup$ I suggest that you try checking the Handbook of Number Theory (currently in two volumes) to see if your conjectured equation is there. Other than that, I do not remember having seen this relationship before. $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 24 '19 at 5:52
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Yes, this relationship is true and can be proved as follows (FYI $\sigma_1(n)=\sum_{d\mid n}d$ is more commonly known as the sum of divisors function, while the divisor function tends to refer to $\sum_{d\mid n}1$).

$$ \sum_{1\leq n\leq x}\frac{\sigma_1(n)}{n}M(x/n) = \sum_{1\leq n\leq x}\frac{1}{n}\sum_{d\mid n}d\sum_{1\leq m\leq x/n}\mu(m). $$

The right-hand side is

$$ = \sum_{1\leq d,m\leq x} d\mu(m)\sum_{\substack{1\leq n\leq x/m\\ d\mid n}} \frac{1}{n} = \sum_{1\leq dmn\leq x} \frac{\mu(m)}{n}= \sum_{1\leq k\leq x}\mu \ast 1\ast f(k)$$

where $f(t)=1/t$ and $\star$ denotes the multiplicative Dirichlet convolution. By Mobius inversion $\mu\star 1$ is the identity function for convolution, and hence this sum is just $\sum_{1\leq k\leq x}f(k)=H(x)$ as you claim.

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